Question

此代码有效，但我觉得必须有更好的方法，而不必使用Array.find()两次。

＆＃13;

const people = [
  { id: 0, age: 99 },
  { id: 1, age: 54 },
  { id: 2, age: 54 }
];
const roles = [
  { pId: 0, responsabilites: ['make money'] },
  { pId: 1, responsabilites: ['make money', 'complain'] },
  { pId: 4, responsabilites: ['make the world a better place', 'sarcasmm'] },
];

let roomsAndOrders = people.filter(p => {
  return roles.find(r => r.pId === p.id);
});
roomsAndOrders = roomsAndOrders.map(p => {
  let r = roles.find(r => r.pId === p.id);
  return { ...r, ...p };
});
   
console.log(roomsAndOrders);

＆＃13;

Answer 1

只需使用一个.map。您的原始filter没有多大意义 - filter过滤掉您不想要的数组中的元素，但它不会更改元素。您正在从filter函数返回对象，并且对象是真实的，因此filter实际上并没有做任何事情。

修改：或者只是反过来映射 - 一次从roles映射到people，而不是从people映射到roles。

＆＃13;

const people = [
  { id: 0, age: 99 },
  { id: 1, age: 54 },
  { id: 2, age: 54 }
];
const roles = [
  { pId: 0, responsabilites: ['make money'] },
  { pId: 1, responsabilites: ['make money', 'complain'] },
];

const roomsAndOrders = roles.map(role => {
  const person = people.find(({ id }) => role.pId === id);
  return { ...role, ...person };
});
console.log(roomsAndOrders);

＆＃13;

要仅包含ID在两个数组中的对象，您必须使用.reduce，因为map始终返回与原始数组中相同数量的元素：

＆＃13;

const people = [
  { id: 0, age: 99 },
  { id: 1, age: 54 },
  { id: 2, age: 54 }
];
const roles = [
  { pId: 0, responsabilites: ['make money'] },
  { pId: 1, responsabilites: ['make money', 'complain'] },
  { pId: 4, responsabilites: ['make the world a better place', 'sarcasmm'] },
];

const roomsAndOrders = roles.reduce((accum, role) => {
  const person = people.find(({ id }) => role.pId === id);
  if (person) accum.push({ ...role, ...person });
  return accum;
}, []);
console.log(roomsAndOrders);

＆＃13;

Answer 2

您可以使用哈希表在O（n）中执行此操作：

 const result = [], hash = {};

 for(const person of people)
   result.push(hash[person.id] = {...person});

for(const role of roles)
  if(hash[role.pId])
    Object.assign(hash[role.pId], role);

如何在一个循环中组合/合并/交叉两个对象数组？

2 个答案: